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From Wikipedia:

Given a topological vector space $(X,τ)$ over a field $F$, $S$ is

called bounded iffor every neighborhood $N$ of the zero vectorthere

exists a scalar $α$ so that $$

S \subseteq \alpha N $$ with $$

\alpha N := \{ \alpha x \mid x \in N\}. $$

I was wondering if the concept is still the same when “for every neighborhood $N$ of the zero vector” is replaced by “there exists a neighborhood $N$ of the zero vector”? Is it true that every neighborhood of the zero vector can be scaled to contain any other neighborhood of the zero vector?

- Part (b) of Exercise 13 of first chapter of Rudin's book “Functional Analysis”
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Thanks and regards!

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No. For example, as Mariano commented, $X$ is a neigbhorhood of the zero vector.

If $N$ is a neighborhood of the zero vector such that every neighborhood of the zero vector can be scaled to contain $N$, then $N$ is bounded. It is true that $S$ is bounded if there exists a *bounded* neighborhood $N$ of the zero vector and a scalar $\alpha$ such that $S\subseteq \alpha N$, but this is not equivalent to boundedness in general (See Matt E’s comment below).

E.g., think of $\mathbb R^2$, where boundedness is the same as being contained in a big enough circle. A set like $\{(x,y):x^2+y^2<1\}$ cannot be scaled to contain a set like $\{(x,y):x>-1\}$. More generally, in a normed space boundedness of $S$ is equivalent to $\sup\{\|x\|:x\in S\}<\infty$, and the idea is that $S$ can be scaled to be contained in an open neighborhood $N$ of the zero vector, *no matter how small $N$ is*. (Of course in general $X$ need not even be metrizable, so this intuition isn’t perfect.)

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